Homework 6 Solutions

4 m smaller than sun mass but not atmosphere wont

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Unformatted text preview: ed gas! n = 104 m ­3 µ = 0.5 (1 mH per 2 particles) => MJeans = 1011 M¤ => a star definitely cannot form straight out of the ISM (c) [5 pts] The Sun’s photosphere (hint: look back at prior homeworks for estimate of n) T = 5800 K n ≈ 1022 m ­3 µ = 1 (mostly neutral H) => MJeans = 0.4 M¤ => smaller than Sun mass but not atmosphere – won’t work (d) [5 pts] The core of the Sun T = 1.5x107 K µ = 4/3 (mostly ionized He => 4 mH per 3 particles) ρcore ≈ 150 g/cm3 = 1.5x106 kg/m3 ncore ≈ ρcore/µmH = 7x1032 m ­3 => MJeans = 0.1 M¤ => hmm, this seems promising! Unfortunately the part of the sun that is current fusing is smaller than this. But it indicates a point where gravity might win – and stellar evolution starts it’s next phase… (e) [5 pts] The Earth’s atmosphere (assume standard temperature, pressure, and gas composition near sea level) T = 293 K, P = 105 N/m2 (both STP) µ = 0.78 µN2 + 0.21 µO2 + 0.01 µAr = 0.78 (28) + 0.21 (32) + 0.01 (40) = 29 n = P/kT = 2x1025 m ­3 => MJeans = 1.2x10 ­7 M¤ = 0.04 Earth masses – hmm, this also seems promising, but if you do the numbers Earth’s atmosphere is 10 ­6 Earth masses, so we don’t have to worry about atmosphere stars (but what about Jupiter…?) (2) [55 pts] Circumstellar disks The mass density of an axisymmetric circumstellar disk is commonly parameterized as a function of radius r and vertical scaleheight z in the following manner: ρ(r, z ) = ρ(r, 0)e − z2 H (r ) 2 Σ(r) = Σ0 (r/R0 )α H (r) = H0 (r/R0 )β where H(r) is the vertical scaleheight of the disk as a function of radius, Σ(r) is the column density o...
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This document was uploaded on 02/26/2014 for the course PHYS 160 at UCSD.

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